AP Calculus AB 7.6 Finding General Solutions Using Separation of Variables - MCQs - Exam Style Questions
No-Calc Question
Let \(y=f(x)\) be the particular solution to the differential equation \(\dfrac{dy}{dx}=(1+y)\cos x\) with initial condition \(f(0)=2\). What is the value of \(f\!\left(\dfrac{\pi}{6}\right)\) ?
(A) \(\sqrt{e}-1\)
(B) \(\sqrt{e}+1\)
(C) \(2\sqrt{e}\)
(D) \(3\sqrt{e}-1\)
▶️ Answer/Explanation
Separate variables: \(\dfrac{dy}{1+y}=\cos x\,dx\).
Integrate: \(\ln|1+y|=\sin x+C\).
Use \(f(0)=2\): \(\ln 3=C\Rightarrow 1+y=3e^{\sin x}\).
Thus \(f(x)=3e^{\sin x}-1\).
Evaluate at \(x=\dfrac{\pi}{6}\): \(f\!\left(\dfrac{\pi}{6}\right)=3e^{1/2}-1=3\sqrt{e}-1\).
✅ Answer: (D)
No-Calc Question
If \(\dfrac{dy}{dx}=2-y\), and if \(y=1\) when \(x=1\), then \(y=\)
(A) \(2-e^{x-1}\)
(B) \(2-e^{\,1-x}\)
(C) \(2-e^{-x}\)
(D) \(2+e^{-x}\)
(B) \(2-e^{\,1-x}\)
(C) \(2-e^{-x}\)
(D) \(2+e^{-x}\)
▶️ Answer/Explanation
Separate: \(\dfrac{dy}{2-y}=dx\).
Integrate: \(-\ln|2-y|=x+C\Rightarrow 2-y=Ke^{-x}\).
Use \(y(1)=1\): \(2-1=Ke^{-1}\Rightarrow K=e\).
Hence \(y=2-e^{\,1-x}\).
✅ Answer: (B)